Solve ∫ (from 0 to 3) of (-16+4.33x)(-4+0.67x) wrt x

$$\int_{ 0 }^{ 3 } (-16+4.33x)(-4+0.67x) d x$$

Evaluate

$$91.9299$$

Solution Steps

Apply the distributive property by multiplying each term of -16+4.33x by each term of -4+0.67x.

$$\int _{0}^{3}64-10.72x-17.32x+2.9011x^{2}\mathrm{d}x$$

Combine -10.72x and -17.32x to get -28.04x.

$$\int _{0}^{3}64-28.04x+2.9011x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 64-\frac{701x}{25}+\frac{29011x^{2}}{10000}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64\mathrm{d}x+\int -\frac{701x}{25}\mathrm{d}x+\int \frac{29011x^{2}}{10000}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 64\mathrm{d}x-\frac{701\int x\mathrm{d}x}{25}+\frac{29011\int x^{2}\mathrm{d}x}{10000}$$

Find the integral of 64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$64x-\frac{701\int x\mathrm{d}x}{25}+\frac{29011\int x^{2}\mathrm{d}x}{10000}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -28.04 times \frac{x^{2}}{2}.

$$64x-\frac{701x^{2}}{50}+\frac{29011\int x^{2}\mathrm{d}x}{10000}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2.9011 times \frac{x^{3}}{3}.

$$64x-\frac{701x^{2}}{50}+\frac{29011x^{3}}{30000}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$64\times 3-\frac{701}{50}\times 3^{2}+\frac{29011}{30000}\times 3^{3}-\left(64\times 0-\frac{701}{50}\times 0^{2}+\frac{29011}{30000}\times 0^{3}\right)$$

Simplify.

$$\frac{919299}{10000}$$